Set the properties of an auto-generated ContentPresenter

Question

Consider the following part of a visual tree in a Windows Phone SL application:

As you can see, a PanoramaItem template contains a reference to a static DataTemplateSelector. It is a simple class that dynamically feeds the data template based on a provided Key, to display different views for different view models provided as DataContext to PanoramaItem. Here is the code for it:

public static T FindResource<T>(this DependencyObject initial, string key) where T : DependencyObject
        {
            DependencyObject current = initial;

            while (current != null)
            {
                if (current is FrameworkElement)
                {
                    if ((current as FrameworkElement).Resources.Contains(key))
                    {
                        return (T)(current as FrameworkElement).Resources[key];
                    }
                }

                current = VisualTreeHelper.GetParent(current);
            }

            if (Application.Current.Resources.Contains(key))
            {
                return (T)Application.Current.Resources[key];
            }

            return default(T);
        }
    }

    public class DataTemplateSelector : ContentControl
    {
        protected override void OnContentChanged(object oldContent, object newContent)
        {
            ContentTemplate = this.FindResource<DataTemplate>(newContent.GetType().FullName);
        }
    }

The problem is that I have no control over the creation of ContentPresenter you can see selected on an image above. To get a consistent layout, I need to be able to set it's Vertical Alignment property. I don't seem to know how I can do it, since I can't reference this ContentPresenter. How can I set a ContentPresenter's properties?

Answer 1

The solution yet again was simple:

Define a style for my ContentControl derived class:

<Style TargetType="support:DataTemplateSelector">
            <Setter Property="Template">
                <Setter.Value>
                    <ControlTemplate TargetType="support:DataTemplateSelector">
                        <ContentPresenter
                        ContentTemplate="{TemplateBinding support:DataTemplateSelector.ContentTemplate}"
                        Content="{TemplateBinding support:DataTemplateSelector.Content}" 
                        VerticalAlignment="Top"/>
                    </ControlTemplate>
                </Setter.Value>
            </Setter>
        </Style>

I have defined mine in UserControl.Resources section of a View's XAML.

The call the "restyling" line of code in the class's constructor:

 public class DataTemplateSelector : ContentControl
    {
        public DataTemplateSelector()
        {
            this.DefaultStyleKey = typeof (DataTemplateSelector);
        } 

And that's how you can control the ContentPresenter element's look of a ContentControl's derived control.

Answer 2

You seem to already have the answer sitting there in front of you.

Use the VisualTreeHelper.GetParent to get the direct visual parent of the DataTemplateSelector as a FrameworkElement, then modify its VerticalAlignment property.

BTW :-

            if (current is FrameworkElement) 
            { 
                if ((current as FrameworkElement).Resources.Contains(key)) 
                { 
                    return (T)(current as FrameworkElement).Resources[key]; 
                } 
            } 

can be reduced to:-

            var currentFE = current As FrameworkElement;
            if (currentFE != null && currentFE.Contains(key) )
            {
                return (T)currentFE.Resource[key];
            }

Cast attempt only occurs once instead of potentially three times, is shorter and is easier to understand. An example of where adding a variable is beneficial (whereas normally we might try to reduce the number of variables in our code).